SQL查询avg差异显示错误？

Question

试图找出两个avg给出错误之间的区别？

http://sqlfiddle.com/#!3/7160d/9

select * from
(
select avg(avg_stars) as avg_1
from
(
select r.mid, avg(stars) as avg_stars
from
rating r inner join movie m
on r.mid = m.mid
where year < '1980'
group by r.mid
)
)
-
(
select avg(avg_stars) as avg_2
from
(
select r.mid, avg(stars) as avg_stars
from
rating r inner join movie m
on r.mid = m.mid
where year > '1980'
group by r.mid
)
)

Answer 1

应该是“一点点”简化

select 
avg(case when year < '1980' then stars end) -
avg(case when year > '1980' then stars end)
from movie m
inner join
(select mId, Cast(stars as int) as stars from Rating) r
on m.mID = r.mID

或（如果星号类型正确，则删除演员表）

select 
avg(case when year < '1980' then cast (stars as int) end) -
avg(case when year > '1980' then cast(stars as int) end)
from movie m
inner join
Rating r
on m.mID = r.mID

请参阅SqlFiddle

Answer 2

这是你想要的。

select 
(
select avg(avg_stars) as avg_1
from
(
select r.mid, avg(convert(int,stars)) as avg_stars
from
rating r inner join movie m
on r.mid = m.mid
where year < '1980'
group by r.mid
) x
) 
-

(
select avg(avg_stars) as avg_2
from
(
select r.mid, avg(convert(int,stars)) as avg_stars
from
rating r inner join movie m
on r.mid = m.mid
where year > '1980'
group by r.mid
) x
) 

Answer 3

另一种方法：

select [stats].Period, avg([stats].stars) as avg_stars    
from 
(
  select
    stars, 
    (case when year > '1980' then 'after 1980' else 'before 1980' end) as [Period]  
  from rating r inner join movie m on r.mid = m.mid
) [stats]
group by [stats].Period

（如果您date为year，则可以将句点设为date

http://sqlfiddle.com/#!3/2bf4e/26